# Minimum representation of total generalized variation

I need help understanding the proof of Theorem 3.1 from this paper. To make this post self-containded, at first I briefly review the definitions.

Let $$\Omega \subset\mathbb{R}^d$$ be a bounded domain and $$\alpha =(\alpha_0,\, \alpha_1) > 0$$. The functional assigning each $$u \in L_1(\Omega)$$ the value $$\begin{equation*} \begin{split} \text{TGV}^2_{\alpha}(u) = \sup\, \bigg\{\int_{\Omega} u \, \text{div}^2 v\, dx : \, v\in C_c^2 (\Omega, S^{d\times d}),\,&|| v ||_{\infty} \leq \alpha_0,\, \\ &|| \text{div}\, v||_{\infty} \leq \alpha_1 \bigg\} \end{split}\end{equation*}$$ is called the Total Generalized Variation of second order of $$u$$.

Here, $$S^{d \times d}$$ is the set of symmetric matrices, $$C_c^2 (\Omega, S^{d\times d})$$ is the vector space of compactly supported, twice continuously differentiable $$S^{d \times d}$$-valued mappings and $$\text{div}\, v \in C_c^1 (\Omega, \mathbb{R}^{d})$$ and $$\text{div}^2 v \in C_c(\Omega)$$ are defined by $$(\text{div}\, v)_i = \sum_{j=1}^{d} \frac{\partial v_{ij}}{\partial x_j}, \quad \text{div}^2v = \sum_{i=1}^{d} \frac{\partial^2 v_{ii}}{\partial x_i^2} + 2\sum_{i

Theorem. Suppose that $$\text{BD}(\Omega)$$ denotes the space of vector fields of Bounded Deformation, i.e., $$w\in L^1(\Omega, \mathbb{R}^d)$$ such that the distributional symmetrized derivative $$\mathcal{E}w = \frac{1}{2}(\nabla w + \nabla w^T)$$ is an $$S^{d \times d}$$-valued Radon measure. Then $$\text{TGV}_{\alpha} ^2 (u)= \min_{w \in \text{BD}(\Omega)} \alpha_1 ||\nabla u -w ||_{\mathcal{M}} + \alpha_0 || \mathcal{E}w ||_{\mathcal{M}}.$$

Proof. Choosing $$X = C_0^2 (\Omega, S^{d\times d})$$, $$Y = C_0^1 (\Omega, \mathbb{R}^{d})$$, $$\Lambda = \text{div} \in \mathcal{L}(X,Y)$$ and for $$v\in x$$, $$w\in Y$$, $$F_1(v) = \iota_{||\cdot||_{\infty}\leq \alpha_0} (v),$$ $$F_2(w) = \iota_{||\cdot||_{\infty}\leq \alpha_1}(w) - \int_{\Omega} u \, \text{div} w \, dx.$$ By a simple substitution we observe that $$\text{TGV}_{\alpha}^2 (u) = -\inf_{v\in X} F_1(v) + F_2(\Lambda v).$$ Furthermore, $$Y=^?\bigcup_{\lambda \geq 0} \lambda \big( \text{dom}(F_1) - \Lambda \text{dom}(F_2)\big)$$ hence by Fenchel-Rockafellar duality it follows that $$\text{TGV}_{\alpha}^2(u)=\min_{w\in Y^*} F_1^*(-\Lambda^*w ) + F_2^*(w) .$$

Now, $$Y^* = C_0^1(\Omega, \mathbb{R}^d)^*$$ can be regarded as a space of distributions and the dual functionals can be written as $$\begin{equation*} \begin{split} F_1^*(-\Lambda^*w) &=^? \begin{cases} \alpha_0 || \mathcal{E}w ||_{\mathcal{M}} & \text{if }w\in\text{BD}(\Omega)\\ \infty & \text{else}, \end{cases} \\ F_2^*(w) &=^? \begin{cases} \alpha_1||\nabla u - w||_{\mathcal{M}} & \text{if } \nabla u -w \in \mathcal{M}(\Omega, \mathbb{R}^d)\\ \infty & \text{else}. \end{cases} \end{split} \end{equation*}$$ Since $$\text{BD}(\Omega) \subset \mathcal{M}(\Omega, \mathbb{R}^d)$$, the result follows. $$\square$$

I need help with the three question-marked equalities. For the first one, fidgeting in my chair, I can convince myself why $$\supseteq$$ holds but for the converse inclusion I'm totally blank.

As for the dual functionals, although it has not been mentioned in the paper, $$F_1^*$$ and $$F_2^*$$ are most likely meant to be convex conjugates of the corresponding functionals which are support functionals here, but I can't just figure out why these equalities hold.

Also what exactly is $$w^T$$ in the statement of the theorem, and how are we allowed to add its gradient to $$\nabla w$$? Should they be viewd as matrices? Are they even of the same dimension?

Your help is much needed and appreciated!